Given sequence is √2 , √8, √18,…
a1 = √2 ,
a2 = √8 = √ ( 2 × 2 ) ×2 = 2√2
a3 = √18 = √ ( 3 × 3 ) × 2 = 3 √2
First term (a ) = a1 = √2
a2 – a1 = 2√2 – √2 = √2
a3 – a2 = 3√2 – 2√2 = √2
Therefore ,
a2 – a1 = a3 – a2 = √2
Common difference (d) = √2
The given sequence is in A.P .
nth term = an = a + ( n-1 ) d
14th term of the A.P = a4
a4 = a + ( 4 – 1 ) d
= a + 3d
= √2 + 3 × √18
= √2 + 3√2
= 4√2
4th term of the given A.P = 4√2
or
a4 = 4√2